Question

what maximum volume of 3 M solution of KOH can be prepared from 1L each of 1 M KOH and 6 M KOH solution by using water​

Answer 1

Answer:

Maximum volume of 3 M solution of KOH can be prepared from 1 L of 1 M KOH solution is 0.3333 L.

Maximum volume of 3 M solution of KOH can be prepared from 1 L of 6 M KOH solution is 2 L.

Explanation:

1) Molarity of the KOH solution = $M_1=1 M$

Volume of KOH solution = $V_1=1 L$

Molarity of required KOH solution = $M_2=3 M$

Maximum Volume of KOH solution = $V_2=?$

$M_1V_1=M_2V_2$ (Dilution)

$1M\times 1L=3 M\times V_2$

$V_2=0.3333 L$

2) Molarity of the KOH solution = $M_1=6 M$

Volume of KOH solution = $V_1=1 L$

Molarity of required KOH solution = $M_2=3 M$

Maximum Volume of KOH solution = $V_2=?$

$M_1V_1=M_2V_2$ (Dilution)

$6 M\times 1L=3 M\times V_2$

$V_2=2 L$

Answer 2

Answer:

Maximum volume of 3 M solution of KOH can be prepared from 1 L of 1 M KOH solution is 0.3333 L.

Maximum volume of 3 M solution of KOH can be prepared from 1 L of 6 M KOH solution is 2 L.

Explanation:

1) Molarity of the KOH solution = M_1=1 MM

1

=1M

Volume of KOH solution = V_1=1 LV

1

=1L

Molarity of required KOH solution = M_2=3 MM

2

=3M

Maximum Volume of KOH solution = V_2=?V

2

=?

M_1V_1=M_2V_2M

1

V

1

=M

2

V

2

(Dilution)

1M\times 1L=3 M\times V_21M×1L=3M×V

2

V_2=0.3333 LV

2

=0.3333L

2) Molarity of the KOH solution = M_1=6 MM

1

=6M

Volume of KOH solution = V_1=1 LV

1

=1L

Molarity of required KOH solution = M_2=3 MM

2

=3M

Maximum Volume of KOH solution = V_2=?V

2

=?

M_1V_1=M_2V_2M

1

V

1

=M

2

V

2

(Dilution)

6 M\times 1L=3 M\times V_26M×1L=3M×V

2

V_2=2 LV

2

=2L

So, total volume = 2 + 0.33

= 2.33L